Integral Test

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Theorem

Let $f$ be a real function which is continuous, positive and decreasing on the interval $\left[{1 \,.\,.\, +\infty}\right)$.

Let the sequence $\left \langle {\Delta_n} \right \rangle$ be defined as:

$\displaystyle \Delta_n = \sum_{k \mathop = 1}^n f \left({k}\right) - \int_1^n f \left({x}\right) \rd x$


Then $\left \langle {\Delta_n} \right \rangle$ is decreasing and bounded below by zero.

Hence it converges.


Proof

From Upper and Lower Bounds of Integral, we have that:

$\displaystyle m \left({b - a}\right) \le \int_a^b f \left({x}\right) \rd x \le M \left({b - a}\right)$

where:

$M$ is the maximum

and:

$m$ is the minimum

of $f \left({x}\right)$ on $\left[{a \,.\,.\, b}\right]$.

Since $f$ decreases, $M = f \left({a}\right)$ and $m = f \left({b}\right)$.

Thus it follows that:

$\displaystyle \forall k \in \N_{>0}: f \left({k + 1}\right) \le \int_k^{k + 1} f \left({x}\right) \rd x \le f \left({k}\right)$

as $\left({k + 1}\right) - k = 1$.


Thus:

\(\displaystyle \Delta_{n + 1} - \Delta_n\) \(=\) \(\displaystyle \left({\sum_{k \mathop = 1}^{n + 1} f \left({k}\right) - \int_1^{n + 1} f \left({x}\right) \rd x}\right) - \left({\sum_{k \mathop = 1}^n f \left({k}\right) - \int_1^n f \left({x}\right) \rd x}\right)\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle f \left({n + 1}\right) - \int_n^{n + 1} f \left({x}\right) \rd x\) $\quad$ $\quad$
\(\displaystyle \) \(\le\) \(\displaystyle f \left({n + 1}\right) - f \left({n + 1}\right)\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle 0\) $\quad$ $\quad$


Thus $\left \langle {\Delta_n} \right \rangle$ is decreasing.

$\Box$


Also:

\(\displaystyle \Delta_n\) \(=\) \(\displaystyle \sum_{k \mathop = 1}^n f \left({k}\right) - \sum_{k \mathop = 1}^{n-1} \int_k^{k+1} f \left({x}\right) \ \mathrm dx\) $\quad$ $\quad$
\(\displaystyle \) \(\ge\) \(\displaystyle \sum_{k \mathop = 1}^n f \left({k}\right) - \sum_{k \mathop = 1}^{n-1} f \left({k}\right)\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle f \left({n}\right)\) $\quad$ $\quad$
\(\displaystyle \) \(\ge\) \(\displaystyle 0\) $\quad$ $\quad$

$\Box$


Hence the result.

$\blacksquare$


Also known as

The integral test is also known as the Euler-Maclaurin Summation Formula, but that result properly refer to a more precise theorem of which this is a simple corollary.


Notes

It follows from this that if $f$ is continuous, positive and decreasing on $\left[{1 \,.\,.\, \infty}\right)$, then the series $\displaystyle \sum_{k \mathop = 1}^\infty f \left({k}\right)$ and the improper integral $\displaystyle \int_1^{\mathop \to +\infty} f \left({x}\right) \rd x$ either both converge or both diverge.

So this theorem provides a test for the convergence of both a series and an improper integral.


Sources